Simplify the following expression: $y = \dfrac{2x^2- 3x- 27}{2x - 9}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(-27)} &=& -54 \\ {a} + {b} &=& &=& {-3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-54$ and add them together. Remember, since $-54$ is negative, one of the factors must be negative. The factors that add up to ${-3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-9}$ and ${b}$ is ${6}$ $ \begin{eqnarray} {ab} &=& ({-9})({6}) &=& -54 \\ {a} + {b} &=& {-9} + {6} &=& -3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-9}x) + ({6}x {-27}) $ Factor out the common factors: $ x(2x - 9) + 3(2x - 9)$ Now factor out $(2x - 9)$ $ (2x - 9)(x + 3)$ The original expression can therefore be written: $ \dfrac{(2x - 9)(x + 3)}{2x - 9}$ We are dividing by $2x - 9$ , so $2x - 9 \neq 0$ Therefore, $x \neq \frac{9}{2}$ This leaves us with $x + 3; x \neq \frac{9}{2}$.